Car Pooling

Description

There is a car with capacity empty seats. The vehicle only drives east (i.e., it cannot turn around and drive west).

You are given the integer capacity and an array trips where trips[i] = [numPassengersi, fromi, toi] indicates that the ith trip has numPassengersi passengers and the locations to pick them up and drop them off are fromi and toi respectively. The locations are given as the number of kilometers due east from the car’s initial location.

Return true if it is possible to pick up and drop off all passengers for all the given trips, or false otherwise.

Example 1:

Input: trips = [[2,1,5],[3,3,7]], capacity = 4 Output: false

Example 2:

Input: trips = [[2,1,5],[3,3,7]], capacity = 5 Output: true

Constraints:

• 1 <= trips.length <= 1000
• trips[i].length == 3
• 1 <= numPassengersi <= 100
• 0 <= fromi < toi <= 1000
• 1 <= capacity <= 105

Code

同 My Calendar III。

Time Complexity: $O(n\log n)$, Space Complexity: $O(n)$

class Solution {
public:
    map<int, int> mp;
    bool carPooling(vector<vector<int>>& trips, int capacity) {
        
        for(auto& trip: trips) {
            mp[trip[1]] += trip[0];
            mp[trip[2]] -= trip[0];
        }
 
        int passenger = 0;
        int maxPassenger = 0;
        for(auto it = mp.begin(); it != mp.end(); it++) {
            passenger += it->second;
            if(passenger > capacity) return false;
        }
 
        return true;
 
    }
};

其實不需要用到 map，因為不需要考慮要不要插入，只需要檢查，因此可以只用 array。

Time Complexity: $O(n)$, Space Complexity: $O(n)$

class Solution {
public:
    bool carPooling(vector<vector<int>>& trips, int capacity) {
        
        int passengers[1001] = {};
 
        for(auto& trip: trips) {
            passengers[trip[1]] += trip[0];
            passengers[trip[2]] -= trip[0];
        }
 
        int curPassengers = 0;
        for(int i = 0; i < 1001; i++) {
            curPassengers += passengers[i];
            if(curPassengers > capacity) return false;
        }
 
        return true;
 
    }
};

Source

• Car Pooling - LeetCode